poj1065&1548 dp,最长上升子序列,偏序与反偏序

题目描述:
  给定n个数对(xi,yi),对这些数对进行划分,保证划分出来的每一堆都是一条单调的链,问最少可以划分成多少堆
  原题如下:
Wooden Sticks
Time Limit: 1000MS   Memory Limit: 10000K
Total Submissions: 26409   Accepted: 11423

Description

There is a pile of n wooden sticks. The length and weight of each stick are known in advance. The sticks are to be processed by a woodworking machine in one by one fashion. It needs some time, called setup time, for the machine to prepare processing a stick. The setup times are associated with cleaning operations and changing tools and shapes in the machine. The setup times of the woodworking machine are given as follows: 
(a) The setup time for the first wooden stick is 1 minute. 
(b) Right after processing a stick of length l and weight w , the machine will need no setup time for a stick of length l' and weight w' if l <= l' and w <= w'. Otherwise, it will need 1 minute for setup. 
You are to find the minimum setup time to process a given pile of n wooden sticks. For example, if you have five sticks whose pairs of length and weight are ( 9 , 4 ) , ( 2 , 5 ) , ( 1 , 2 ) , ( 5 , 3 ) , and ( 4 , 1 ) , then the minimum setup time should be 2 minutes since there is a sequence of pairs ( 4 , 1 ) , ( 5 , 3 ) , ( 9 , 4 ) , ( 1 , 2 ) , ( 2 , 5 ) .

Input

The input consists of T test cases. The number of test cases (T) is given in the first line of the input file. Each test case consists of two lines: The first line has an integer n , 1 <= n <= 5000 , that represents the number of wooden sticks in the test case, and the second line contains 2n positive integers l1 , w1 , l2 , w2 ,..., ln , wn , each of magnitude at most 10000 , where li and wi are the length and weight of the i th wooden stick, respectively. The 2n integers are delimited by one or more spaces.

Output

The output should contain the minimum setup time in minutes, one per line.

Sample Input

3 
5 
4 9 5 2 2 1 3 5 1 4 
3 
2 2 1 1 2 2 
3 
1 3 2 2 3 1 

Sample Output

2
1
3

Source

poj1548原理相同,不放了
  首先我们要求单调链,把样例中的点画在坐标系中,我们便可看出,要求出单调链,要对数据排序(二维降一维)
  poj讨论区有较为清楚的描述(本人理解但是无法用语言描述)
第一次贪心:其实并非每次都要求一个最长的序列,然后删除,只要选出的序列符合"极大"即可。这里的极大和函数的极值是一个概念。
例如,(1,4) (4,4) (5,3) (9,4) 其中最大(最长)的一个序列为 (1,4)->(4,4)->(9,4)
极大的序列 为 (5,3)->(9,4)
证明:很简单,因为所有的木头都需要被加工,并且与加工的先后顺序无关,那么该木头一定
需要在某个序列中,而我们每次选择的序列是极大的,故每次操作都没有遗漏可加入的木头,因此总的操作数是最少的。
 
  然后我们就要开始求单调链的数量,这里用到一个定理:(这个定理我不知道啊。。。QAQ)
  对于Dilworth定理,给出如下解释:
  一种Dilworth定理的等价表述是:在有穷偏序集中,任何反链最大元素数目等于任何将集合到链的划分中链的最小数目。一个关于无限偏序集的理论指出,在此种情况下,一个偏序集具有有限的宽度w,当且仅当它可以划分为最少w条链。
  可知,对于已经排序好的数据,上升单调链条数=下降单调链的最大长度(任何反链最大元素数目等于任何将集合到链的划分中链的最小数目)
  所以这道题最终转化成了求最长下降子序列的问题
  代码如下:
  
 1 #include<iostream>
 2 #include<cstdio>
 3 #include<algorithm>
 4 using namespace std;
 5 struct node
 6 {
 7     int l,w;
 8 }wood[10000];
 9 bool cmp(node a,node b)
10 {
11     if(a.l==b.l) return a.w<b.w;
12     return a.l<b.l;
13 }
14 int dp[10000];
15 int T,n;
16 void solve()
17 {
18     for(int i=1;i<=n;i++)
19     {
20         dp[i]=1;
21         for(int j=1;j<=i;j++)
22         {
23             if(wood[j].w>wood[i].w)
24                 dp[i]=max(dp[i],dp[j]+1);
25         }
26         //printf("now:%d
",dp[i]);
27     }
28 }
29 int main()
30 {
31     //freopen("in.txt","r",stdin);
32     cin>>T;
33     for(int i=1;i<=T;i++)
34     {
35         cin>>n;
36         for(int j=1;j<=n;j++)
37         {
38             scanf("%d %d",&wood[j].l,&wood[j].w);
39         }
40         sort(wood+1,wood+1+n,cmp);
41         //for(int j=1;j<=n;j++) printf("%d %d
",wood[j].l,wood[j].w);
42         solve();int ans=0;
43         for(int j=1;j<=n;j++) ans=max(ans,dp[j]);
44         printf("%d
",ans);
45     }
46     return 0;
47 }
View Code
原文地址:https://www.cnblogs.com/codeoosacm/p/10028326.html